QA 

081 



LIBRARY OF CONGRESS. 



Shelf..£>:&/J 

UNITED STATES OF AMERICA. 



L 



DIFFERENTIATION OF FUNCTIONS. 



NOTES 



DIFFEKENTIATION OP FUNCTIONS 



WITH EXAMPLES. 



/ 



I. Functions 1 

II. Differential Coefficient 3 

III. Differentiation of Functions 6 

IV. Successive Differentiation 34 

V. Differentials . 38 



PRINTED, NOT PUBLISHED. 



Copyright by George A. Osborne. 




J. S. Gushing & Co., Printers, 115 High Street, Boston. 



I. FUNCTIONS. 

1. When two variable quantities are so related that the value 
of one depends upon that of the other, the former is said to be 
a function of the latter. Thus, the volume of a sphere is a 
function of its radius. Any expression containing x is a func- 
tion of x, as ar 2 , log(# +CC 2 ), sin 3#, etc. 

2. The symbols F(x), f(x), <£(#), <K#)> etc., are used to 
denote functions of x. The following examples will illustrate 
the notation of functions. 

EXAMPLES. 

1. If cf>(x) = x 2 + 5, show that 

<£(2) = 9, <K1) = 6, <K0) = 5, <Ka+l)=a 2 + 2a+6, 
cj>(x + 7i) = x 2 + 27ix + 7i 2 + 5. 

2. If /(a?) = 2x z - of - 7a + 6, show that 

/(3) = 30, /(2) = 4, /(0)=6, /(1) = 0, 
/(-2) = 0, /(f) = 0, f(x-2) = 2a*-13a* + 21x, 
f(x + 7i)=2x s +(6h-l)x 2 + (6h 2 -27i-7)x + 27i s 

3. Given <j> (x) = x 2 — 6 2 , ^(#) = x + a ; show that 

-M = a- 6 , *£+*>=„. 

4. Given /i(y)~2*«-j/»+l, f 2 (y) = 7y 2 - 6y+l ; show that 

/i(l)'=/ 2 (l), /i(f)=/ 2 (|), /!(-2)=/ 2 (-2), 
/i<0)=/.(0). 



5. If f(x) = , show that 

J v } l+x 

f 2 (x) = x, where f 2 (x) denotes /[/(a?)]. 

6. If cf> (m) = (m + 1) m [m — 1) (ra — 2) , show that 

*(2)=*(1) = *(0)=*(-1) = 0, *(8) = *(-2), 

<£(m+l) <f>(m) 
m + 2 ~ m — 2 



7. If (a;) = (a; — a) (a; — b) (x — c) , show that 
4>(a + b) • <l>(b + c) ■ 4>(c + a) 



C0(O)]* 



: -8^> 



a + 5 + c\ 



*(2«) <M25) *(2c) =4(aS+ j, + ^_ 3(a6+6c+co)j 

a o c 

♦(-)-*(-»>-*(-') _, Wa+t+ . )] .. 



II. DIFFERENTIAL COEFFICIENT. 

3. In the equation y = ar 2 , if we suppose x to vary, y will 
vary also. To fix the attention upon a definite value of a?, let 
us suppose x = 10 and therefore ;?/ =100, and let us inquire 
what addition or increment will be produced in y by a certain 
increment assigned to x. 

Let Ax denote the increment of #, and Ay the increment 
of y, and in general the symbol A denotes an increment in 
the quantity following it. 

Now, x being 10, if we calculate the value of Ay resulting 
from different values of A#, we find results as in the following 
table : — 



If A* = 


then A y = 


Ay 
Ax 


3. 


69. 


23. 


2. 


44. 


22. 


1. 


21. 


21. 


0.1 


2.01 


20.1 


0.01 


0.2001 


20.01 


0.001 


0.020001 


20.001 



The third column gives the value of the ratio between the 
increments of x and of y. 

It appears from the table that, as A# diminishes and 
approaches zero, Ay also diminishes and approaches zero; 

A v 

—?- diminishes, but instead of approaching zero, approaches 20 

Za X 

as its limit. 



This limit of — ^ is denoted by -^, and is called the differ- 
A# dx 

ential coefficient of y with respect to x. So, in this case, when 

x = 10, ^ = 20. 
dx 



4. Without restricting ourselves to any one numerical value, 
we may obtain _? from the equation y = x? thus : — 

Having y = x 2 , let A# = 7i, and let the new value of y be 

denoted by 

y = (a? + A) 2 ; 

therefore, 

Ay = y' — y = (» + ft) 2 — a? = 2a* + 7* 2 . 

Dividing by A# = 7i, gives 

A# 
The limit of this, when h approaches zero, is 2x. Hence, 

dx 



5. In the same way the differential coefficient of any other 
given function may be found. 

In general, if y = <f> (x) , 

Ay = y'— # = £(« + A) -£(a?), 

Ay = <ft(s + ft) — <ft(s) 

Aa; 7i 

$?= limit of *(* + *)-*(»), 
a# 7*. 

as ft approaches zero. 



The differential coefficient of a function may then be defined 
as the limiting value of the ratio of the increment of the func- 
tion to the increment of the variable, as these increments 
approach zero. 

EXAMPLES. 

Find the differential coefficients in the f ollowino; : — 



& ■ 



dy 



1. y = 3x i -2x. -^-=6a?-2. 
u dx 

2. y = x* + h. lH 8 ^- 

3. y = (x-l)(2x + 3). || = 4* + 1. 

A 1 

4 ' y = x 

- a 

x — a 

6. v = • 

y x + a 

7. y = ^/x. 



8. y=V^ = ~2. 
2 



9. y = 

-Vx + 1 

10. y=-fyx. 



dy = 

dx 


1 


dy__ 
dx 


2a 


dy 


2a 


dx 


(x + a) 2 


dy _ 


1 


dx 


2^/x 


dy __ 


X 


dx 


V^-2 


dy 


1 


dx 


0*+l)« 


dy _ 


1 


dx 


3a;f 



III. DIFFERENTIATION OF FUNCTIONS. 

6. The process of finding the differential coefficient of a 
given function is called differentiation. The examples on 
page 5 are intended to explain more fully the nature of the 
differential coefficient. Differentiation may, however, be most 
readily performed by the application of certain general rules 
which may be expressed by formulae. 

In these formulae the letters w, v, w, denote variable quanti- 
ties, which may be functions of x ; and c and w, constant 
quantities. 

7. The following are the formulae for 



ALGEBRAIC FUNCTIONS. 

^ = (1) 

dx 

d_, x _du.dv ,a\ 

dx dx dx 

A( M + c) = ^ (3) 

dx dx 

d , x du , dv //( x 

— (uv) = v h u — (4) 

dx dx dx 

d , v du , K \ 

Yx {CU)=ZC Tx ( } 

\JLJs IAjJs 

du dv 

u — 

d fu\ _ dx dx , fi . 

dx \vj v 2 

±(u-)=nu^^ (7) 

dx dx 



8. The following special cases of the preceding formulae 
will be useful : — 

A(M±0±t0±etc.) = — ± — ± — ±etc. (2') 
dx dx dx dx 

d f x du . dv , dio /Af s 

— (uvw)=vw \-uw [-uv — • • • (4 r ) 

dx dx dx dx 



dx 


(") 


dx 

c 


d 


v^ 


die 

dx 


dx 


2^/u 



(6') 



• • • - ■ (7') 

We proceed to give the derivation of these formulae. 

9. Formula (1). This follows from the nature of a con- 
stant, which can undergo no change. Hence 

A c = and — = ; 

Ax 

therefore, its limit 

^ = 0. 
dx 

10. Formula (2). Let y = u + v, and suppose that when x 
is changed into x + h, y, u, and v become y\ u\ and v' ; then 

y f = u' + v f ; 
therefore 

y* — y = u' — u + v f — v ; 
that is, 

A?/ = Aw + Av. 

Divide by A a?; then 

Ay__Aw , Av 
Ax Ax Ax 



8 

Now suppose Ax to diminish and approach zero, and we 
have, for the limits of these fractions, 
dy __du . dv 
dx dx dx 

If in this we substitute for y, u + v, we have formula (2). 

do 
11. Formula (3). This is a special case of (2), — being 

dx 
zero. 



12. Formula (4). Let y = uv] then, as in (2), 
y 1 = v! r o\ 

y f — y = u f V* — UV = (u 1 — U)V f + U (v* — v) , 



that is, 

Ay = v'Au + uAv. 

Divide by A a?; then 

Ay ,Au : Av 

Ax Ax Ax 

Now suppose A& to approach zero, and, noticing that the 

limit of v ! is v, we have 

dy du , dv 

-£ = v \-u — • 

dx dx dx 

dc 
13. Formula (5). This is a special case of (4), — being 

dx 

zero. But we may derive it independently thus : — 
y—cu, 



y' =z cu\ 




y' — y = c(u'' 


-u), 


Ay = cAw, 




Ay __ Au 
Ax Ax 




dy ___ du 
dx dx 





14. Formula (6). Let 2/ = -, 



then 
therefore 

that is, 



, u 1 



. u* u u'v — uv* (u , — u)v—u(v , —-v) 

V V V V V V 



A v A u — ^£ A v 

&y = - f j 

Aw Av 

^ V u 

Ay Ax Ax 

Ax" v'v % 

Now suppose Ace to diminish towards zero, and, noticing 
that the limit of v ! is i>, we have 

du dv 

V u 

dy . dx dx 

dx~~ v 2 

Or we may derive (6) from (4) thus : — 
Since 



therefore 

By (4), 



It 

V 



yv = u. 



therefore 



dy . dv du 

v— +y — = — , 

dx dx dx 

dy__du u dv , 
dx dx v dx ' 



du dv 

V u 

dy __ dx dx 

dx" v 2 



10 



15. Formula (7). First, suppose n to be a positive integer. 

Let 

y =u n , 

and, as before, 

y' = u'% 

y* — y = i(' n — u n 

= (u 1 — u) (it'"- 1 + u tn ~ 2 u + u ,n - B u 2 ~- u n ~ l ) , 

that is, 

A y = A u (u'"- 1 + u' n - 2 u + u tn - s u 2 --- n"' 1 ) , 

*y = o'*- 1 + u ' n - 2 u + u ,n - s u 2 ... w"- 1 ) — . 

Ace A# • 

Now let A x diminish ; then, u being the limit of u\ each 
of the n terms within the parenthesis becomes w*" 1 ; therefore 

— 2- = 7i?^ n x — . 
dx dx 

P 
Second, suppose n to be a positive fraction, -• 

Let 



y = w, 

then 

y* = u* ; 

therefore 

But we have already shown (7) to be true when the exponent 
is a positive integer ; hence we may apply it to each member 
of this equation. This gives 

qy q Y -f- = pu p *— ; 
ax dx 

therefore 

dy __p u*' 1 du 
dx~~qy <l - l dx 



11 

s 
Substituting for y, w«, gives 

dy __p p_ x du 
dx~~ q u dx 

which shows (7) to be true in this case also. Hence that 
formula applies to any positive value of n, whether integral 
or fractional. 

Third, suppose n to be negative and equal to — m. 
Let 

by (6), 

-SL(u m ) _ m ^-i^f 
dy dx dx , du 

dx u 2m u 2m "" dx 

Hence (7) is universally true. 

16. Formula (2') is an extension of (2), and may be derived 
in the same way. 



Let 



that is, 



Hence 



y = u±v ± w + etc., 
y } = u' ± v' ± w' + etc., 
y f — y = u f — u± (V — v) ± (w 1 — w) + etc. ; 

Ay = Au ± Av ± Aw + etc., 

A?/ Au , Av , Aw , 
—£ = — ± — ± h etc. 

Ax Ax Ax Ax 



dy^du idv + dWi * 
dx dx dx dx 



12 

17. Formula (4 f ) is an extension of (4), and may be derived 
from it thus : — 

d , v d , x . dm 

— (uv 'IV) =w — (uv) + uv — 
dx dx dx 

( du . dv\ . dw 

= w[v \-u — )+uv — 

\ dx dx) dx 

du . dv . dw 

= vw f- uw 1- uv 

dx dx dx 

Similarly, for the product of four functions, we have 

d , N du , dv , dw . dz 

— (uvwz) = vwz h wzu f- zuv h uvw — 

dx dx dx dx dx 

The formula may be extended to the product of any number 
of functions. 

dc 

18. Formula (6 f ) is easily derived from (6), — being zero. 

ax 

19. Formula (7 f ) is an application of (7) to the square root 
of a function. 

du 

dx dx dx 2^/u 



EXAMPLES. 

1. y = x* + x 2 . ^- = Sx 2 + 2x. 

2. y = 3aJ*-4a?+6oj 8 -12a?+5. ^ = 12 (g?-a* + x - 1). 

dx 

x 5 x* dy 4 x 3 

* 5 12 da? 3 

4. y=6a£ + 10a>*-ar*. ^= 10a* + 6arf + ^3. 

cZ# 3 



13 



5. y=(a? + 2)t. ^=s5*(as* + 2)*. 

(XX 

dy 10 a? 



3. y = Voa? 4 -— 4. 



*» vW_ 4 



7. y={x+iy(2x^-iy. ^=(ie x+ i)( X+ iy(2x-iy. 

(XX 

a + bx + ex 2 ch/ a 

* x ax x- 

J x \ Ox 3 ^3 

10 — a; + ^ % 3 — 6 a — a? 

" y ~a? + Z do; - (a^ + 3) 2 ' 



%i + x — x? +a dy _2xi — x-\-2x% — 3a 



11. 2/ = 

£2 



<te 2a?t 



12. Given 

(a + a;) 5 = a 5 + o a 4 a? + 10a 3 x 2 + 10a 2 x 3 + 5 ax 4 + x? ; 
derive by differentiation the expansion of (a + x)*. 

13. Given 1+s + s 2 — +af t = a?l>+1 "" 1 ; 

a? — 1 

derive the sum of the series 1 + 2 a? + 3 a? 2 • •• + ?ia? n_1 . 

rfy_ 1 



14. y=J±*. 
\1 — » 



da? 



(l-aOVl-aJ 2 



15. y= ** . *. 



(1+a?) 71 da? (l+a?) n+1 

16. ^ = (l-2a?+3a? 2 -4a? 3 )(l+a?) 2 . ^ = -20a? 3 (1 + a?). 

clx 

17. - 2/ = (l-3ir 2 + 6a; 4 )(l+£c 2 ) 3 . ^ = 60« 5 (1+^) 2 . 

clx 



14 



18. y = rf(a + Sxy(a-2xy 



^ = 5x*(a + 3a;) 2 (a - 2x) (a 2 + 2ax - 12 a 2 ). 

CLX 



19. y = x 15 (a-3x) 5 (a + 5x) 3 . 



^==15x 1 \a--3xy(a + 5xy(a 2 +2ax-23x i ). 
dx 



20. y = (a + x) m (b + xy 



c ly==[in(b+x)+n(a+x)] (a+x) m -\b+x) n -\ 
cix 



21. y = 



22. 2/ = 

23. y = 



(a + x) m (b+x) n 

dy __ m (b + x) + n(a + x) 
dx" (a + x) m+1 (b + x) n+1 ' 

x dy __ 1 



-Vl-x 2 

1—x 

VT+a?' 



dx (l-a?)* 
d?/ _ 1 + a? 



do; 



(l + ar 9 )t 



24. y 



3+ar 2 ' 



dx (3+a?) 2 ^/x 



25. 2/ = 



1 



dy __ x_ 



x + Vl + a 2 



26. 2/ = 



*» VT+a 2 

^_ L 



1. 



27. 2/ = 



aj+ Vl-a 2 

■y/cT+x + Va — a? 



dx 2a;(l-a 2 )+Vl-i 



Va + # t Va — cc 



^y _ _ a2 + a Va 2 — tf 2 
da; 



28. 2/ = 



3ar> + 2 



^y. 



ft 2 Va 2 — ar 2 
2 



a?(a? + l)t 

29. 2/ = 3(a3 2 + l)*(4a^~3). 



da? ar 9 (a^+l)i 

^=56a^(ar 2 +l)i 
dx 



15 

oq __ V(# + a) 3 dy __ (x — 2 a) Va? + a 

■yjx — a dx (x — a)i 

3l vrp?+vri^? t d y= 2/ n i 

VT+^?-Vl-^ dx x s \ Vl-a 4 

32. yr-f * V- ^- 52- 



l + Vl-xV *» Wl-a; 2 



LOGARITHMIC AND EXPONENTIAL FUNCTIONS. 

20. The formulae for differentiation are as follows : — 

d , Mdu /ox 

— log a ^ = -— - (8) - 

dx u ax 

where M = log a e = 

log e a 

d , 1 du /nx 

— logu = * (9) 

dx u dx 

— a* = loga.a«— (10) 

dx B dx v J 

A e u ==e u — (11) 

dx dx 

A u * = vu v - 1 — + logu-if— (12) 

dx dx dx 

21. Before deriving these formulas it is necessary to find 
the limit of the expression 

M-U-j, as z approaches infinity. 
V */ 

By the Binomial Theorem 
which ma}' be written 

* The base of the logarithm is e, unless otherwise expressed. 



16 



Now when z increases indefinitely, we have 

I* 15 

This quantity is usually denoted by e, so that 



limit of fl + -\=l + l + etc. 



e=l+i-f r^+r-+etc. 

1 [2 [3 

The value of e can be easily calculated to any desired 
number of decimals by computing the values of the successive 
terms of this series. For seven decimal places the calculation 

is as follows, — 

1. 
1. 

.5 

.166666667 

.041666667 

.008333333 

.001388889 

.000198413 

.000024802 

.000002756 

.000000276 

.000000025 

.000000002 



e = 2. 7182818-. 

By calculating the value of f 1 + - j for different; values of z, 
we may verify its limit. Thus 

(l+i) 2 =2.25 

(1 + i) 5 =2.48832 

(1 _|_ _i_) io = 2.59374 

(1.01) 100 = 2.70481 

(1.001 ) 1000 = 2.71692 

(1.0001) 10000 = 2.71815 

(1.00001) 100000 = 2.71827 

(1.000001) 1000000 = 2.71828 



17 



22. Formula (8). Let y =log a u, 

then 

y' =log a (u + Au), 

u-\- Au 



Ay = log a (u + Aw)- log a u == log a - 



16 



/ A?A Aw, / Au\±u 
= log (l+-) = -log a (l+-) • 

Dividing by Ax, 

A v / AnV M 1 Aw 

— = loo* 1 A 

Ax 1U ^\^ u) U Ax 

Now if A# approach zero, Aw at the same time approaches 

u 

zero ; then the limit of ( 1 + - — ] " is the same as the limit 

of ( 1 + -) as z increases indefinitely. But we have already 

found the latter limit to be e. Hence we have 
dy \og a e du 
dx ~~ u dx 

23. Formula (9) is a special case of (8) when a = e. In 

this case 

31= log e e = 1. 

24. Formula (10) may be derived from (9) as follows, — 
Let y = a u ; 

taking the logarithm of each member, we have 

logy = u log a ; 
therefore by (9) 

1 dy du # 

y dx~ ° dx ' 

multiplying by ?/ = a M , we have 

dy _ dw 

— = log a • a w j — 
dx ° da; 



18 

25. Formula (11) is a special case of (10) when a = e. 

26. Formula (12) . Let y = u v ; taking the logarithm of each 
member, we have 

log y = v log u ; 

therefore by (9) 

1 dy v du dv u 

y dx ~~ u dx ° dx ' 

multiplying by ?/ = u v , we have 

dv ,dz£ _ dv 

-f- = v^- 1 — - + log u • u v ~— 
dx dx ° da; 



1. ?/ = log(3a^ + a;). 

2. 2/ = a: log x. 

3. ?/ = afloga;. 



EXAMPLES. 

dy _ 6rc+l 



4. y = log Vl — x 2 . 

5. 2/ = e x (l — x 3 ). 

6. 2/ = V^ - log(V a? + 1 )- 

7. y = log(loga?). 

8. y = log(^ + e-). 



do; 


3 x 2 + a; 




d?/ _ 
da; 


: 1 +loga?. 




dx 


: af*" 1 (l + 71 lo 


ga>). 


dv 


05 




dx 


1-ar 9 




dy _ 
dx 


:e*(l — 3 a?- 


a; 3 ). 


dy 


1 




dx 


2(V* + 1) 




dy __ 


1 




dx 


a; log a; 




dy _ 


e 2x -l 





da; e 2x + 1 



19 

9. y = (x-3)e 2 * + 4:xe x + x. '^■ = (2x-o)e ix +4(x+l)e x +l. 

dx 

10. y = log w {5x + x 3 ). ^ = jf|±M, 

da; 5x + x 3 

where Jf = — — r - = log 10 e = .434294 
log e 10 

11. 2/ = 5^ + 2 ^ ||=2(z + l)5* 2 + 2 *log5, 

log 5 = 1.609440 

12. y = ^-^- *~ * 



e x + e~ x dx (e x + e~ x ) 2 

What is the result of differentiating both members of each 
of the three following equations? 



5V ' 2 3 4 

-4ns. = 1— x + x 2 — x? + 

1+x 

14. logl±?=2(» + J+^ + ^+. 

1 — # V 3 5 7 

,4**. — — - = l+;c 2 +a; 4 + a; 6 + ... 

1-ar 2 . 

/y»2 /)«3 /y»4 

15. e*=l + a; + f- + f- + f-+... 

-4ms. e x =l+a; + -^ + ^H 

[2 [3 

16. y = x n a x . ^- = x n - l a x {n + x\oga). 

ClX 

iff i / c%\ 4(#— 1) dii a^ + 4 

17. y = log(x— 2) * 4-. -^== — -. 

* 6V } (x-2) 2 dx (x-2) B 

is. y-iog V^+v* ^- v a 

-^/a — y'a; da; (a — a?) y'a; 



20 



19. y 

20. y 

21. y 

22. y 

23. y 

24. 2/ 

25. y 

26. y 

27. 2/ 

28. y 

29. ?/ 
SO. 2/ 

31. 2/ 

32. 2/ = | 



a; logo; 



flog(l-a;). 



dy logic 



1-a; ' &v y da; (1-a;) 2 

: e -A(a;l — 3a; + 6a;i — 6) . 



dx 



:| 4 [(loga;) 2 -logV^ + i]. jg = a?(loga;) 2 



o X , u 0? b 



a a 4 a' 



da; 



:loga;.log(loga;)-loga;. dy ^ogQogx) ^ 

dx x 

dy 



:log(o?-3 + Vo? 2 -6a?+13). g£= x 

(to Vaj 2 --6a;+13 

: mlog (y'a? + Vo? + m) + Vma; + a? 2 . 

^ _. |m + a; 
d® \ a? 



:l0g 
:l0g 
:10ff 



X 



dy_ 



a — Va 2 — a? 2 
x V2 + Vl + i 



dx 
dy 



x Va 2 — x 2 

V? 



Vl-a? 2 

Va-g + q 2 + Va? 2 + ft 2 
Va^ + a 2 - yV + 6 2 ^ Va? 2 + a 2 Va? 2 + 6 2 



dx (l-^Vl+a 8 



, la?— 1 . fa^ + l 



dy __ x 2 — 1 
do? o? 4 + ^+ 1 



: fa"— e-¥ ( e 2x + 2e 4 *+3e 6 *). ^= 24 e 6 * (e 2 * - 1). 

dx 



i 



— = X * (l-logo?). 

^ = .pYYi + io g ? 

da? \nj V ^ 



21 

33. y = (ex) x . ^ = (ex)*(2 +log»). 

dx 

35. ?/ = aj 10 ^. ^^logo 2 .^*" 1 . 

dx 

36. v = ^^ ^=0. 
* ' dx 

37. 2/ = e e *. ^L=e* x e°. 

dx 

38. ?/ = e* x . ^ = e* x af (1+loga?). 

39. y = x* x . ■ ^=2/^[J + logo; + (log a) 2 ! 



TRIGONOMETRIC FUNCTIONS. 
27. The formulae for differentiation are as follows : — 

— smu = cosu — (13) 

dx dx 

d du / . ,v 

— cosw = — smw — (14) 

dx dx 

— tanw = sec - ^ — (lo) 

dx dx 

— cot^ = — cosec 2 ^ — (16) 

dx dx 

— seeu=secut&nu— . (17) 

dx dx 

— cosecM = — coseewcotw — (18) 

dx dx 



22 



— vers 26 = sin w — (19) 

do; c?# 

— covers u = — cos zt — (20) 



28. Formula (13) . Let ?/ = sinw, 

then 

y ] = sin(w + Ate), 

therefore 



A y = s>in(u + Au) — sinu 

2 A it 

cos [u-\ - ) sin — • 






Hence *i n ^ u 

sin ■ 

Ay f . Au\ 2 

_ ^ = cos it H . 

Ax V 2 Aw 



Now when Ax approaches zero, Au likewise approaches 

. Au 

sin 

2 

zero, and the limit of is unity ; 

' Au J ' 

2 

therefore 

dy du 

-£ = cos w — . 
dx dx 



29. Formula (14) may be derived by substituting in (13) 

for u, u. 

2 



The * d . 

— sin 
dx 



[ u) = COS W ) — [ uu 

V dx) 



or 

d . / du\ . cZm 

— cos u = smu[ = — sm u — 

dx V dx dx 



23 



sin u 

30. Formula (15). Since tam£ = , 

v J cos u 

d . d 

cos ^ — sin u — sin i£ — cos u 

by (o) — tan u ■■ 

dx 



cos 2 


u 




2 C?W . . 2 

cosn6 h sir 

dx 


du 

u — 

dx 


du 

dx 


COS 2 16 




' cos 2 u 


9 cfa^ 
= sec^j — 
dx 







31. Formula (16) may be derived from (15) by substituting 
- — u for u in the same way that (14) was derived from (13) . 



32. Formula (17). Since secw = - 



COSty 



d . du 

cos u sin u — 

, //1X d dx dx 

b y ( 6 ) - T -secu = - 2 = s — 

du 

= sec u tan u — 
dx 



33. Formula (18) may be derived from (17) by substituting 
2 



v — w for ^ 



34. Formula} (19) and (20) are readily obtained from (13) 
and (14) by the relations 

vers w= 1 — cosw, 
covers u = 1 — sin it. 



24 



EXAMPLES. 

1. y = sin 2 x cos x. — = 2cos2xcosx — sin2xsinx. 

dx 

2. ?/ = tan 2 5x. -2= 10 tan 5 x sec 2 5 x. 

3. ?/ = tanx — x. — - = tairx. 

dx 

4. y = sin (wx + m) . — = ?? cos (?ix + m) . 

cix 

K tanx— 1 dy 

o. y = — = sinx4-cosx. 

sec x dx 

6. y = sin 3 x cosx. -^ = sin 2 x(3 cos 2 x — sin 2 x) . 

dx v J 

7. ?/ = sin(x + a) cos(x — a).-^ = cos2x. 

dx 

sin (a — x) dy . Q 0/ . N 

8. i/ = * -« -^- = — sm2acosecr(a + x). 

sin (a + x) dx 

9. ?/ = tan 2 x — log(sec 2 x). -^=2tan s x. 

10. y = tan 4 x — 2tan 2 x + log(sec 4 x). 

^ = 4tan 5 x. 

11. ?/ = (asin 2 x + 5cos 2 x) n . 

-^ = n(a — 6) sin2x(a sin 2 x + b cos 2 x) n " 1 . 

dx v } K } 

12. ?/ = lo2:smx. -^=cotx. 
* dx 

13. y = log tan x. 



dx sin 2 x 



14. y = losf sec x. -^ = tan x. 

* dx 



25 



15. y = vers(- + s jversf- 



7T 

S 

2 



-£ = — sin 2 s. 
dx 



16. v = g-0* sins -cobs). ^/ = e «* s ins/ 
* a 2 + 1 ds 

i« -in. ^V /sins . \ 

17. y = a«« ^ = y^_ + coga .io ga? j. 

18. y = sin?is sin n s. -^ = ?i sin*^ 1 ^ sin(?i +1) s. 

ds 

-q _ sin m ns d?/ _ mn sin m " 1 ^scos(m — -n) x 

cos n mx dx 

dy_ § 



dx 1 + tans 



20. y = x + log cos ( x — - j. 

21. y — logtanf -+ - j. -^=secs. 
* 6 12 4>/ ds 



oo i 1 — coss c?y 

53^. 2/ = log A -^ = cosec s. 

\ 1 + cos s dx 

oo i /a cos x — b sins cfa/ — ab 

\acoss + 6sins dx a 2 cos 2 s— -6 2 sin 2 s 

0/ i _ tans — tan 3 s dy . 

#4. y = -^- = cos4s. 

sec 4 s dx 



In each of the following pairs of equations derive by differen- 
tiation each of the two equations from the other : — 

25. sin 2s = 2 sins coss, 
cos 2s = cos 2 s — sin 2 s. 

oc • o 2 tans 

26. sin 2 s = — , 

1 + tan-s 

rt 1 — tan 2 s 

cos 2 s = — • 

l + tan J s 



26 

27. sin 3 x = 3 sin x — 4 sin 3 #, 
cos3a; = 4cos 3 # — 3 cos a?. 

28. sin4# = 4 sin a; cos 3 x — 4 cos a? sin 3 #, 
cos 4 a; = 1 — 8 sin 2 # cos 2 x. 

29. sin (m + n) x = sin mx cos tie + cos mx sin ?ioj, 
cos (m -\-n)x = cos wmb cos ?i# — sin mx sin ?ix 

oj' ' o»5 /v,7 

30. sin a? = a; — - — hr i — h ••• 

-_. — p. — 

I? I* L^ 

Q1 . e x^l_ e -x^Tl 

31. sm# = === — , 

2V-1 

cos a; = ■ 



INVERSE TRIGONOMETRIC FUNCTIONS. 
35. The formulae for differentiation are as follows : — 

— sin l u = — — (21) 

dx Vl — u 2 dx 

d _i 1 du / M x 

— cos i w = === — (22) 

d# Vl — v? dx 

Atan- 1 ^— — 9 — (23) 

±cot" 1 u = — ? — (24) 

dx 1 + w 2 cfo v 

— sec 1 m = - — ..*••• (2o) 

cfe u Vw 2 — 1 da? 



27 



d _i 1 du / OC . 

— cosec l u = — (20) 

dx w Vw 2 -l^ 

— vers I tt= — — - (2/) 

do; V2 a — u 2 dx 



36. Formula (21). Let # = sirr 1 w, 

therefore 

smy = u; 



by (13) 

therefore 



cosy— = • 
dx dx 



but 

therefore 



dy _ 1 du t 

dx cosy dx 

cosy = Vl — sin 2 y = Vl — u\ 

dy __ 1 du 

dx -y/i—ytdx 



37. Formula (22) may be derived by a similar method to 
that used for (21). 

38. Formula (23). Let y = tan~ 1 w, 

therefore 

tau y = u; 



by (15) 


o dy du 
se&y-^-= — 5 

dx dx 


therefore 






dy _ 1 du u 




dx sec 2 y dx ' 


but 






sec 2 y = 1 + tan 2 y = 1 -f u 2 , 


therefore 






dy _ 1 du 




dx 1 + ir dx 



28 

39. Formula (24) may be derived by a similar method to 
that used for (23). 

40. Formula (25). Let y = sec" 1 !*, 

therefore 

sec?/ = u ; 

sec y tan y-£ = — 
dx dx 

therefore 

c7?/ __ 1 du t 

cZ# sec?/ tan?/ cto 



but 

therefore 



sec?/ tan?/= sec?/ Vsec 2 ?/ — 1 = m Vm 2 — 1, 
c7?/__ 1 <fa& 



41. Formula (26) may be derived in a similar manner. 

42. Formula (27). Let ?/ = vers~ 1 t^ 

therefore 

u = vers ?/ = 1 — cosy ; 

by (14) or (19) 

du . du 
— = smy-^-f 
dx dx 

therefore 

dy __ 1 du u 

dx sin?/ dx 
but 



sin y = Vl— cos 2 ?/ = Vl— (1 — w) 2 = V2 w — w 2 , 

therefore 

cfa/ __ 1 dw 

da? V2 u - u 2 &t 



29 



EXAMPLES. 



1. y = tan" 1 mx. 

2. y = sin- 1 (Sx-l). 



o -l oX 

3. y = vers l — • 
u 9 



4. # = sin" 1 (3 05 — 40?*). 



5 . y^tanij— 



6. ysstairV. 

7. 2/ = tau~ 1 (7itano;). 



3 
8. v = cosec~ 1 — • 
B 2x 



9. 2/ = vers~ i 2aj 2 . 



ia -l 2ar> 

10. w = vers l -• 

* 1 + a,* 2 

11. 2/ = tan- 1 ^=^. 

* 2 

12. ^cosec- 1 ^-^ 

13. 2/ = sec- 1 ^±i. 

* x 2 -! da? a; 2 +l 



da; 


m 
l + wra? 2 


dy _ 


3 


dx 


V6a;-9ar 2 


dy _ 


2 

: • 


da; 


V9a;-4a; 2 


<^_ 


3 


da; 


Vl-a^ 


<fy_ 


2 


da; 


1+a; 2 


d#_ 


1 


da; 


e x + er x 


dy _ 


n 


da; 


cos 2 a; + ?i 2 sm 2 a; 


% _ 


2 


da; 


V9-4a; 2 


d#_ 


2 


da? 


Vl-ar 2 


d?/__ 


2 


dx 


1+a; 2 


dy _ 


2 


da; 


e x + <r x 


^y_ 


2 


da; 


Vl~ & 


%_ 


-2 



30 



14. y 

15. y 

16. y 

17. y 

18. y 

19. y 

20. y 

21. 2/ 

22. y 

23. y 

24. 3/ 

25. 2/ 

26. y 

27. 2/ 



r Sill J — ! . 

V2 

.tan- 1 4sina! 
3+5 cos 



x 



_i3 + 5 
= cos l 

* l Q 



COS a; 



5 + 3 cos a? 
. -i 1 - «? 

= Sill 1 

1+ar 9 

= cosec l — ! — • 
2x 

. _i x + a 

= tan * — ' 

1 — aa; 



= sin" 1 Vsin x. 
= tan-\£= 



\r 



COS 07 



+ COS0J 

1 — Vaa; 



da; 

dx 

da; 5 + 3cosa; 

cZa; 



Vl — 2 a? — x 2 

4 

5 + 8 cos a? 



da; 
da? 



'l+aj 2 
2 



: 1 + ar* 

1 



= l+a? 



-2 = 4Vl + cosec a;, 
da; 



dx 2 

dy _ 
dx 



= cot" 1 - + lo< 



\x — a 

°\x + a 



x \x + a dx 

|(.+VI=?). | = 



_ 1 

~2ya3(l + a0* 

d?/ _ 2 a;r 

da; T a; 4 — a 4 



= tan" 1 

= cos x 

e x + e" z 

= sec X A 

tan" 



Vi- 

efo 2 VI 



da; 



v/i — ar— a; 

^(l + a;Vr^a?) 



f+6~ x * 

d^ = __--l__ 
^ 2 Vl^a? 

(x+a) tan -1 -— ^ax. -*■ = tan -1 -. 
\a da; \a 



31 



28. 


, ,1 + Vl + X 2 
y= cot l — ■ ! 


c/.y _ 1 




dx 2(l+a?) 




29. 


j X tail a 


c??/ __ a 2 tan a 


1 


-ya'—xr 


cfcc a 1 — or -y 
dy _ 1 


' a 2 — a; 2 sec 2 a 


30. 


y= cot * 

5 \2 + x 


' , 




tf» 2V2-a?- 


-X 2 


31. 


?/ = tan I -• 

J 1-3X 2 


C?/7 _ 3 

di 1 + x 2 




32. 


y = tan -1 — — — - + cot" 


1 *' , + taQ- 1 2a. 

1+2.T 2 


dy _ 3 

dx 1+dx 2 


33. 


y = tan~ 1 ' +taa 


Y 2b — x 

xVs 

f tan- 1 2 * • 
1-2& 2 


^ = 0. 
dx 


34. 


5 ^\2a?+2a? + l 


dy 8 s 2 

dx 4# 4 + 1 



(28) 



32 

43. Given — ; to find -2- 

dy dx 

It is evident that A"y__ 1 

Ax~~ Ace' 
Ay 
however small the values of Ao; and Ay. As these quantities 
approach zero, we have, for the limits of the members of this 
equation, 

dy == ± m 

ax (XX 
dy 

44. Given -^ and — ; to find -3L 

dz dx dx 

It is evident that Ay Ay Az 

Ax~~ Az A#' 

however small A a?, Ay, and Az. As these quantities approach 

zero, we have, for the limits of the members of this equation, 

dy_dydz^^ ^ ^v 

dx ~ dz dx 

EXAMPLES. 
In the following eight examples find by differentiation 



— , and then ^ by (28), 
dy dx 



1. x- 



2. x = 



( + 1 

2y 



y-i 



3. x=Vy 2 + l-y. 

a rr~x — = — ^y 2Vl + sin?/ 

dx cosy -y/2 — x 2 



dy (y + l) 2 


a 


dx a 

dy (y-^Y 


X 2 ' 

2 


dx 2 
dy -vV + 1 


(x-2) r 


dX y _ -yjyt _f_ 1 


2X 2 



33 

5. x = taw 1 (y+Vy 2 — 1). j-=2 yVy 2 —1 =%(sec 2 x— cosec 2 x). 

y dy (1 + logij) 2 y 2 

b. x = 



A 



\-\-\ogy dx log?/ xy — x 

e y + -sJ(?y—± dy ^/e 2y — 4 e 2x — 1 

2 da; e 2 ' e 2x +l 

8 a? = 21or A/ey+2+ " v/ ^ := ^ d y= Ve 2y --4 = e 2x -l 
^2 dx e y e 2x + l' 

Compare the two preceding examples with Ex. 8, page 18. 

In the following examples find by differentiation -^ and — ? 
, dz da; 

and then ^ by (29),— 



9. 


y = z 5 , g = a 2 — a; 2 . 




da; 

= -10a?(a 2 -a?*) 4 . 


10 


22 X 
y — jg — 




dy _ 4 




" 32-2' 2 a?.— 1 


da; (a?-2) 2 


11. 


?/ = e z + e 2 % z — log (x — 


x>). 


d 2 = 4o 8 -6aj 8 +l. 
da; 


12. 


y = log (z' s — z) , 2J = e 3x . 




dy __ 5 e 2x — 3 

dx e 2x — 1 


13. 


2/ = log — : — , z = e x . 
z 




d?/ __ e x — e~ x 
dx e x + e~ x 


14. 


y = tan 2 z, 2; = tan" 1 (2ic 


-!)• 


dy _2x 2 -2x + l 
dx 2 (a? — a? 8 ) 2 


15. 




— 2x 
-4a?+l 


dv 1 

dx {x — l) 2 



16 v _, l0 , (« + l)' 1 tan-* 2 '- 1 , -yi+Sx + S^ 
lb. V-^og z ,_ z+1 v _tan v _ , 2- 

dy _ 1 



da; a?2(l-f-a;) 



34 



IV. SUCCESSIVE DIFFERENTIATION. 

« 

45. It has been shown that when y is equal to a given 

function of a?, we may obtain by differentiation the differential 

dv dy 

coefficient, -^- As — will be, in general, also a function of 
dx dx 

#, this may likewise be differentiated, and the result is called 
the second differential coefficient. If this be again differen- 
tiated, the result is called the third differential coefficient ; 
and so on. 



Thus, if 



y = x\ 
dx 
dx dx 

AAiK = 2 ix. 

dx dx dx 



46. The second differential coefficient of y with respect to x 

is denoted by —^- That is, 

J dx 2 



Similarly 



cVy_ : 
dx 2 


d dy 
dx dx 


dx 3 


cl d dy 
dx dx dx 


d*y 


cl cl cl dy 


dx* 


dx dx dx dx 



85 



Or 



Thus if 



d 3 y _ 
dx* 


cfa da; 2 


d 4 y_ 


d d 3 ?/ 


dx* 


da; da? ' 


d n y _ 


d d*" 1 ?^ 


dx n T 


da; dx 11 - 1 


y = 


--x\ 


dy _ 
dx 


-Ax 3 , 


d 2 y_ 

dx 2 


:12a; 2 , 


d 3 y _ 
cfo 3 ~ 


:24a;. 


EXAMPLES. 



1. 2/ = a ; 4 -4ar 3 4-6a; 2 -4a;+l. ^|== 12(a; 2 - 2a; +1). 

2. y = af. ^ = 15. 
* da;° L 

3. y = (x-3)e 2x +4:Xe x -{-x. ^-| = 4e x [(a;-2)e a; +a; + 21. 

dxr 

5. ?/ = a* log x. 

6. ?/ = ar 3 loga;. 

7. y = log(e x +e- x ). 



dx 2 




g.m + 2 




d 2 y_ 
dx 2 


I 

X 






d 4 y_ 


6 






dx 4 


X 






d 3 y_ 


: — 


8(e*— e 


'*) 


dx 3 


(e XJ t-e' 


,)8 



8. y=(a?-6aB+12)e". ^-f 



9. ^Ylogx-f 



10. y = log sin j?. 

11. y = (a^+a 2 ) tan" 1 : 

< 

1:2. ^/ = e" x cos^. 
13. y = tan.r. 



36 




C? 3 ?/ _ 

eta 3 "" 


a?e\ 


d 2 y _ 

dx 2 


xlogx. 


d s y_ 


2 cos # 


dx 3 


sin 3 a? 


d"y _ 


4 a 3 


da? 


(<*+*?)' 


tf 4 ?/ _ 


— 4e _I cos# 



eta 4 

d° ?/ 

— ? = 6 sec 4 # — 4 sec 2 x. 

da? 



14. y: 



,5a?+l, 
x 2 —! 



B=* 



[_(*-!)■' (* + l) 7 . 



Decompose the fraction before differentiating. 



15. ; = ^ Bee 2x. 

16. £ = («(*+ - 



17. 2/ = 



i COS.T COST 35 



27 



do; 2 

c7 2 // 

dx 2 

d?y 

~dx* 



■ Sf-y. 

• n 2 y — ±n(n — l)y n 



= &m°x. 



d 2 y 



18. ii — tan 2 *? + 8 lo^ cos or + 3 x 2 . — \ = 6 tan 4 #. 

dxr . 



19. y = (x 2 -3x + 3)e 2x . 

20. 2/ = ^r3(log.T) 2 -lllogx + — • 



c J-l = 8x 2 e 2x - 
dx 3 

f|=18(logz) 2 . 
oar 



21. v = e cx sin6x. 



i^_2a^ + (tt 2 +& 2 )2/ = 0. 



22. v = sin(wsin- 1 x). (l-a?)^ — x$ + «?y = 0. 

dx- dx 



37 



23. y = acos(log;c) + &sin(loga:)- a?^ + x^+y = 0. 



efcS" 0'4-2)" + 1 ' 
<ta n (3a+4) n+1 ' 



24 


?/ — 




x + 2 


25 


1 

v — 




9 3a? + 4 


26. 


y = e ax . 


27. 


V = log a?. 


9,8 


1-a 

?/ — • 



1 +x 



d n y 
dx n ' 



■ a^e™. 



fy= (-ir|n-l t 
cfo n a? 

d» y== 2(-l)"|n 

da n (l + #) n+1 ' 



2 



Reduce the fraction to a mixed quantity, — 1H — - — , before 

X ~f~" dj 

differentiating. 



29. 



^ a^-4 - dx n ( ' ^|_(aj+2)" +li "(a;-2)» +i J 



ss 



V. DIFFERENTIALS. 

47. The differential coefficient -2 has been defined, not as a 
fraction having a numerator and denominator, but as a single 

bol representing the limiting valne of —?, as A^ and Ay 

A*; 
approach zero. 

Another view of the differential coefficient : _ Is it as an 
1 fraction, dx and dy being defined as infinitely small in- 
crements of x and y, and called differentials of x and y. That 
is, is an infinitely small A as, and dy an infinitely small Ay. 

For instance, if we differentiate y= : . we obtain -^ = 

dx 

Using differentials, ttua result might be written dy = 2 

The first form, -^ — _ asserts that the limit of the ratio of the 
dx 

increment of y to that of x, as these increments approach zero. 
is eqnal :■: : The second form, a = 2xdx, s : t an 

infinitely small increment of y is 2x times the corresponding 
infinitely small increment of x. The two express the same 
relation in different langaage. the first being, howe 
more precise and less open to logical critic:^ . 

These two modes of expression are analogous to the follow- 
::._- ;~ :X^r^:::> ::: :::_■::;; ir^r/v : — 

tfc An infinitely small arc is eqnal to its chord." 

arc 
The limit of the ratio. — — . as these que approach 

zero, is unity 

The first statement may be regarded as an abbreviation of 

the roller and more explicit language of the second. Simi- 

the equation dy = 2xdx may be regarded as a convenient 

substitute for -2 = l 

dx 



39 

48. The formulae for differentiation may be expressed in the 
form of differentials by omitting the dx in each member. Thus, 
formula (4) becomes 

d (uv) = vdu + udv ; 

formula (23), 

, . _j du 

a tan l u- 



1+u 2 ' 
and the others may be similarly expressed. 

Differentiation by the new forms of the formulae is substan- 
tially the same as by the original forms, differing onl} T in using 

the symbol d instead of — • 

J • dx 

For example, take Ex. 10, p. 13. 

dy = d f x + S \ = ( g2 + 3)d(x + 3)-(x + 3) d(s» + 3) 
J \a?+3J (x>+3) 2 

= (x 2 + 3) dx - (x + 3) 2xdx 
"~~ (^ + 3) 2 

= (x 2 +3 — 2x 2 -6x)dx = (3 — ^x — x 2 )dx 
(x 2 + 3) 2 (x 2 + 3) 2 

Dividing by dx gives 

dy ___ 3 — 6x — x 2 

dx~ (x 2 + 3) 2 

49. Successive Differentials. Successive differential coeffl- 

d"V d 3r u 
cients, — ^-, — ^, etc., which have been defined as single sym- 
dx 2 dx? 

bols, may also be interpreted as fractions, the numerators, d 2 y, 
d 3 y, etc., denoting d(dy), d[_d(dy)~\, etc., and called the second, 
third, etc., differentials of y, while the denominators are (dx) 2 , 
(dx) 3 , etc. 



40 

This will be better understood from an example. Let y = x*, 

then 

dy = Ax 3 dx. 

Now, Ax-dx being a variable, dy is a variable, and may 
be again differentiated. Xow, x being the independent vari- 
able, its increment dx may be supposed the same infinitely 
small quantity for all values of x ; that is, we may regard dx as 
a constant in the preceding equation. Thus we obtain 

d (dy) = 12 x 2 dx ■ dx = 12 a?(cto) 8 . 

Denoting d(dy) by d 2 y, 

d 2 y=12x 2 (dx) 2 . 

Differentiating again, and still regarding dx as constant, 

d(d 2 y) = 2±xdx(dx) 2 = 2±(dx)% 
or d s y = 2±(dx) 3 . 

From these equations, by dividing by the power of dx in the 
second members, we find 



{dx) 



2 =1- 
= 24*. 



(dxy 

The variable as, whose differential is supposed constant, is 
called the equicrescent variable. 



